{"id":2518,"date":"2020-12-22T10:31:14","date_gmt":"2020-12-22T05:01:14","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=2518"},"modified":"2020-12-22T12:58:00","modified_gmt":"2020-12-22T07:28:00","slug":"factorization-of-polynomials-using-factor-theorem","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/factorization-of-polynomials-using-factor-theorem\/","title":{"rendered":"Factorization Of Polynomials Using Factor Theorem"},"content":{"rendered":"<h2>Factorization Of Polynomials Using Factor Theorem<\/h2>\n<h3>Factor Theorem:<\/h3>\n<p>If p(x) is a polynomial of degree n \uf0b3 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x).<br \/>\n<strong>Proof:<\/strong> By the Remainder Theorem,<br \/>\np(x) = (x \u2013 a) q(x) + p(a).<br \/>\n(i) If p(a) = 0, then p(x) = (x \u2013 a) q(x), which shows that x \u2013 a is a factor of p(x).<br \/>\n(ii) Since x \u2013 a is a factor of p(x),<br \/>\np(x) = (x \u2013 a) g(x) for same polynomial g(x). In this case, p(a) = (a \u2013 a) g(a) = 0.<\/p>\n<ol>\n<li>Obtain the polynomial p(x).<\/li>\n<li>Obtain the constant term in p(x) and find its all possible factors. For example, in the polynomial<br \/>\nx<sup>4<\/sup> + x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> \u2013 x + 6 the constant term is 6 and its factors are \u00b1 1, \u00b1 2, \u00b1 3, \u00b1 6.<\/li>\n<li>Take one of the factors, say a and replace x by it in the given polynomial. If the polynomial reduces to zero, then (x \u2013 a) is a factor of polynomial.<\/li>\n<li>Obtain the factors equal in no. to the degree of polynomial. Let these are (x\u2013a), (x\u2013b), (x\u2013c.)&#8230;..<\/li>\n<li>Write p(x) = k (x\u2013a) (x\u2013b) (x\u2013c) &#8230;.. where k is constant.<\/li>\n<li>Substitute any value of x other than a,b,c &#8230;&#8230; and find the value of k.<\/li>\n<\/ol>\n<p><strong>People also ask<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/cbselibrary.com\/division-algorithm-for-polynomials\/\">Division algorithm for polynomials<\/a><\/li>\n<li><a href=\"https:\/\/cbselibrary.com\/algebraic-identities-of-polynomials\/\">Algebraic Identities Of Polynomials<\/a><\/li>\n<\/ul>\n<h2>Factorization Of Polynomials Using Factor Theorem Example Problems With Solutions<\/h2>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>Factorize x<sup>2<\/sup>\u00a0+4 + 9 z<sup>2<\/sup> + 4x \u2013 6 xz \u2013 12 z<br \/>\n<strong>Solution:<\/strong><br \/>\nThe presence of the three squares viz.x<sup>2<\/sup>, (2)<sup>2<\/sup>, and (3z)<sup>2<\/sup> gives a clue that identity (vii) could be used. So we write.<br \/>\nA = x<sup>2<\/sup> + (2)<sup>2<\/sup> + (3z)<sup>2<\/sup> + 4x \u2013 6 xz \u2013 12 z<br \/>\nWe note that the last two of the product terms are negative and that both of these contain z. Hence we write A as<br \/>\nA = x<sup>2<\/sup> + (2)<sup>2<\/sup> + (\u20133z)<sup>2<\/sup> + 2.2x \u2013 2.x.(\u20133z) + 2.2 (\u2013 3z)<br \/>\n= (x+2 \u2013 3z)<sup>2<\/sup><br \/>\n= (x + 2 \u2013 3z) (x + 2 \u2013 3z)<\/p>\n<p><strong>Example 2: \u00a0<\/strong> \u00a0Using factor theorem, factorize the polynomial x<sup>3<\/sup> \u2013 6x<sup>2<\/sup>\u00a0+ 11 x \u2013 6.<br \/>\n<strong>Solution:<\/strong><br \/>\nLet f(x) = x<sup>3<\/sup> \u2013 6x<sup>2<\/sup> + 11x \u2013 6<br \/>\nThe constant term in f(x) is equal to \u2013 6 and factors of \u2013 6 are \u00b11, \u00b1 2, \u00b1 3, \u00b1 6.<br \/>\nPutting x = 1 in f(x), we have<br \/>\nf(1) = 1<sup>3<\/sup> \u2013 6 \u00d71<sup>2<\/sup> + 11\u00d7 1\u2013 6<br \/>\n= 1 \u2013 6 + 11\u2013 6 = 0<br \/>\n\u2234 (x\u2013 1) is a factor of f(x)<br \/>\nSimilarly, x \u2013 2 and x \u2013 3 are factors of f(x).<br \/>\nSince f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.<br \/>\nLet f(x) = k (x\u20131) (x\u2013 2) (x \u2013 3). Then,<br \/>\nx<sup>3<\/sup>\u2013 6x<sup>2<\/sup> + 11x \u2013 6 = k(x\u20131) (x\u2013 2) (x\u2013 3)<br \/>\nPutting x = 0 on both sides, we get<br \/>\n\u2013 6 = k (0 \u2013 1) (0 \u2013 2) (0 \u2013 3)<br \/>\n\u21d2 \u2013 6 = \u2013 6 k \u21d2 k = 1<br \/>\nPutting k = 1 in f(x) = k (x\u2013 1) (x\u2013 2) (x\u20133), we get<br \/>\nf(x) = (x\u20131) (x\u2013 2) (x \u2013 3)<br \/>\nHence, x<sup>3<\/sup>\u20136x<sup>2<\/sup> + 11x \u2013 6 = (x\u2013 1) (x \u2013 2) (x\u20133)<\/p>\n<p><strong>Example 3: \u00a0 \u00a0<\/strong>Using factor theorem, factorize the polynomial x<sup>4<\/sup>\u00a0+ x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> \u2013 x + 6.<br \/>\n<strong>Solution:<\/strong><br \/>\nLet f(x) = x<sup>4<\/sup> + x<sup>3<\/sup>\u2013 7x<sup>2<\/sup> \u2013x + 6<br \/>\nthe factors of constant term in f(x) are \u00b11, \u00b12, \u00b13 and \u00b1 6<br \/>\nNow,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-135857\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-1.png\" alt=\"Factorization Of Polynomials Using Factor Theorem 1\" width=\"345\" height=\"399\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-1.png 345w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-1-259x300.png 259w\" sizes=\"(max-width: 345px) 100vw, 345px\" \/><br \/>\nSince f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors<br \/>\nThus, the factors of f (x) are (x\u20131), (x+1),<br \/>\n(x\u20132) and (x+3).<br \/>\nLet f(x) = k (x\u20131) (x+1) (x\u20132) (x + 3)<br \/>\n\u21d2 x<sup>4<\/sup> + x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> \u2013 x + 6<br \/>\n= k (x\u20131) (x +1) (x \u2013 2) (x + 3)<br \/>\nPutting x = 0 on both sides, we get<br \/>\n6 = k (\u20131) (1) (\u20132) (3) \u21d2 6 = 6 k \u21d2 k = 1<br \/>\nSubstituting k = 1 in (i), we get<br \/>\nx<sup>4<\/sup> + x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> \u2013 x + 6 = (x\u20131) (x +1) (x\u20132) (x+3)<\/p>\n<p><strong>Example 4: \u00a0 \u00a0<\/strong>Factorize,\u00a0 2x<sup>4<\/sup>\u00a0+ x<sup>3<\/sup> \u2013 14x<sup>2<\/sup> \u2013 19x \u2013 6<br \/>\n<strong>Solution:<\/strong><br \/>\nLet f(x) = 2x<sup>4<\/sup> + x<sup>3<\/sup> \u2013 14x<sup>2<\/sup> \u2013 19x \u2013 6 be the given polynomial. The factors of the constant term \u2013 6 are \u00b11, \u00b12, \u00b13 and \u00b16, we have,<br \/>\nf(\u20131) = 2(\u20131)<sup>4<\/sup> + (\u20131)<sup>3<\/sup> \u2013 14(\u20131)<sup>2<\/sup> \u2013 19(\u20131)\u2013 6<br \/>\n= 2 \u2013 1 \u2013 14 + 19 \u2013 6 = 21 \u2013 21 = 0<br \/>\nand,<br \/>\nf(\u20132) = 2(\u20132)<sup>4<\/sup> + (\u20132)<sup>3<\/sup> \u2013 14(\u20132)<sup>2<\/sup> \u2013 19(\u20132)\u2013 6<br \/>\n= 32 \u2013 8 \u2013 56 + 38 \u2013 6 = 0<br \/>\nSo, x + 1 and x + 2 are factors of f(x).<br \/>\n\u21d2 (x + 1) (x + 2) is also a factor of f(x)<br \/>\n\u21d2 x<sup>2<\/sup> + 3x + 2 is a factor of f(x)<br \/>\nNow, we divide<br \/>\nf(x) = 2x<sup>4<\/sup> +x<sup>3<\/sup> \u2013 14x<sup>2<\/sup>\u201319x \u2013 6 by<br \/>\nx<sup>2<\/sup> + 3x + 2 to get the other factors.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-135858\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-2.png\" alt=\"Factorization Of Polynomials Using Factor Theorem 2\" width=\"339\" height=\"528\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-2.png 339w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-2-193x300.png 193w\" sizes=\"(max-width: 339px) 100vw, 339px\" \/><\/p>\n<p><strong>Example 5: \u00a0 \u00a0<\/strong>Factorize,\u00a0 9z<sup>3<\/sup>\u00a0\u2013 27z<sup>2<\/sup> \u2013 100 z+ 300, if it is given that (3z+10) is a factor of it.<br \/>\n<strong>Solution:<\/strong><br \/>\nLet us divide 9z<sup>3<\/sup> \u2013 27z<sup>2<\/sup> \u2013 100 z+ 300 by<br \/>\n3z + 10 to get the other factors<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-135860\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-3.png\" alt=\"Factorization Of Polynomials Using Factor Theorem 3\" width=\"335\" height=\"304\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-3.png 335w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-3-300x272.png 300w\" sizes=\"(max-width: 335px) 100vw, 335px\" \/><br \/>\n\u2234 9z<sup>3<\/sup> \u2013 27z<sup>2<\/sup> \u2013 100 z+ 300<br \/>\n= (3z + 10) (3z<sup>2<\/sup>\u201319z + 30)<br \/>\n= (3z + 10) (3z<sup>2<\/sup>\u201310z \u2013 9z + 30)<br \/>\n= (3z + 10) {(3z<sup>2<\/sup>\u201310z) \u2013 (9z \u2013 30)}<br \/>\n= (3z + 10) {z(3z\u201310) \u2013 3(3z\u201310)}<br \/>\n= (3z + 10) (3z\u201310) (z\u20133)<br \/>\nHence, 9z<sup>3<\/sup>\u201327z<sup>2<\/sup>\u2013100z+ 300<br \/>\n= (3z + 10) (3z\u201310) (z\u20133)<\/p>\n<p><strong>Example 6: \u00a0 \u00a0<\/strong>Simplify<br \/>\n\\(\\frac{4x-2}{{{x}^{2}}-x-2}+\\frac{3}{2{{x}^{2}}-7x+6}-\\frac{8x+3}{2{{x}^{2}}-x-3}\\)<br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-135863\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-4.png\" alt=\"Factorization Of Polynomials Using Factor Theorem 4\" width=\"362\" height=\"342\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-4.png 362w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-4-300x283.png 300w\" sizes=\"(max-width: 362px) 100vw, 362px\" \/><\/p>\n<p><strong>Example 7: \u00a0 \u00a0<\/strong>Establish the identity<br \/>\n\\(\\frac{6{{x}^{2}}+11x-8}{3x-2}=\\left( 2x+5 \\right)+\\frac{2}{3x-2}\\)<br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-135864\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/09\/Factorization-Of-Polynomials-Using-Factor-Theorem-5.png\" alt=\"Factorization Of Polynomials Using Factor Theorem 5\" width=\"291\" height=\"221\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Factorization Of Polynomials Using Factor Theorem Factor Theorem: If p(x) is a polynomial of degree n \uf0b3 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x). Proof: By the &#8230; <a title=\"Factorization Of Polynomials Using Factor Theorem\" class=\"read-more\" href=\"https:\/\/cbselibrary.com\/factorization-of-polynomials-using-factor-theorem\/\" aria-label=\"More on Factorization Of Polynomials Using Factor Theorem\">Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[1072,1088,1089,1090,24],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v18.4.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Factorization Of Polynomials Using Factor Theorem - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/factorization-of-polynomials-using-factor-theorem\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Factorization Of Polynomials Using Factor Theorem\" \/>\n<meta property=\"og:description\" content=\"Factorization Of Polynomials Using Factor Theorem Factor Theorem: If p(x) is a polynomial of degree n \uf0b3 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x). 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